HDU1019:Least Common Multiple
2014-08-09 17:54
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[align=left]Problem Description[/align]
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
[align=left]Input[/align]
Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where
m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.
[align=left]Output[/align]
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
[align=left]Sample Input[/align]
2
3 5 7 15
6 4 10296 936 1287 792 1
[align=left]Sample Output[/align]
105
10296
最小公倍数。。
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.
[align=left]Input[/align]
Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where
m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.
[align=left]Output[/align]
For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
[align=left]Sample Input[/align]
2
3 5 7 15
6 4 10296 936 1287 792 1
[align=left]Sample Output[/align]
105
10296
最小公倍数。。
#include<stdio.h> int gcd(int a,int b) { return a%b==0?b:gcd(b,a%b); } int main() { int n,t; scanf("%d",&t); while (t--) { int a=1,b; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&b); a=a/gcd(a,b)*b; } printf("%d\n",a); } return 0; }
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